warning on myqsl_fetch_array

When I conduct a query within a query I get the following warning on myqsl_fetch_array.

I make my first search into one db table. Then taking the result I search the 2nd table, as you see 2 added next to the 2nd search on $con2, $sql2...

<?php
$bookTitle = "";
//echo "<span style='font-weight: bold; font-family: Verdana, Times New Roman, serif; font-size: 24px;'>".$bookTitle."</span><br /><br />\n";
$con = mysql_connect("","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("boti_book", $con);
$i = $_GET['page_num'];
//$keyword = "membership lists";
$sql = "SELECT * FROM boti_css WHERE fontweight LIKE '%bold%'";
$result = mysql_query($sql);
//echo $sql;
while($row = mysql_fetch_array($result))
{
	echo "<span";
	echo " style='font-weight: bold;' ";
	echo "><a href='index.php?page_num=".$row['page_num']."' style='text-decoration: none; color: black;'";
?>
onMouseOver='this.style.color="red"; this.style.textDecoration="underline";' onMouseOut='this.style.color="black"; this.style.textDecoration="none";'
<?php	
	echo ">";
	$searchPage = $row['page_num'].
	$con2 = mysql_connect("","root","");
	if (!$con2)
	  {
	  die('Could not connect: ' . mysql_error());
	  }
	mysql_select_db("boti_book", $con2);
	$i = $_GET['page_num'];
	//$keyword = "membership lists";
	$sql2 = "SELECT * FROM boti_pages WHERE page_num = ".$searchPage;
	$result2 = mysql_query($sql2);// or die(mysql_error());
	//echo $sql;
	while($row2 = mysql_fetch_array($result2))
	{
	echo $row2['content'];
	echo " Page ".$row2['page_num']."</a></span><br /><br />\n";
	}
	//mysql_close($con2);	
}
mysql_close($con);
?>

PHP:

 

$i = $_GET['page_num'];
//$keyword = "membership lists";
$sql = "SELECT * FROM boti_css WHERE fontweight LIKE '%bold%'";
$result = mysql_query($sql);
//echo $sql;
while($row = mysql_fetch_array($result))
{
    echo "<span";
    echo " style='font-weight: bold;' ";
    echo "><a href='index.php?page_num=".$row['page_num']."' style='text-decoration: none; color: black;'";
?>
onMouseOver='this.style.color="red"; this.style.textDecoration="underline";' onMouseOut='this.style.color="black"; this.style.textDecoration="none";'
<?php   
    echo ">";
    $searchPage = $row['page_num'].
    $con2 = mysql_connect("","root","");
    if (!$con2)
      {
      die('Could not connect: ' . mysql_error());
      }
    mysql_select_db("boti_book", $con2);
    $i = $_GET['page_num'];
    //$keyword = "membership lists";
    $sql2 = "SELECT * FROM boti_pages WHERE page_num = ".$searchPage;
    $result2 = mysql_query($sql2);// or die(mysql_error());
    //echo $sql;
this looks like the right line
    while($row2 = mysql_fetch_array($result2))

PHP:

ok so from what i see your problem is with $searchPage = $row['page_num'].
how are you getting $row['page_num'] when you have row set to something with css and $i = $_GET['page_num'];
as the place to get the page num from anyway i dont know what you got in that part of the db but i think you just got that little typeo also have a link with the $row['page_num'] thing too does it work for you